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Trigonometric identities are equations that relate to different trigonometric functions and are true for any value of the variable that is there in the domain. Basically, an identity is an equation that holds true for all the values of the variables present in it. The function of an angle i.e the angles and sides relationships is given by trigonometric functions. Sine, cosine, tangent, cotangent, Cos, Cosec are called the trigonometric functions. Let’s look into the Cos Square theta formula below.

## Formula for Cos Square theta

According to the trigonometric identities, we know that,

cos^{2}θ + sin^{2}θ = 1

where,

- θ is an acute angle of a right triangle.
- sinθ and cosθ are the trigonometric ratios given as follows:

sinθ = Altitude/Hypotenuse

cosθ = Base/Hypotenuse - sin
^{2}θ is the square of sinθ and cos^{2}θ is the square of cosθ i.e,

sin^{2}θ = (sinθ)^{2}

cos^{2}θ= (cosθ)^{2}

Thus cos square theta formula is given by,

cos^{2}θ = 1 – sin^{2}θ

## Things to Remember

- Trigonometric ratios are the value of the trigonometric functions that are based on the ratio of the primary trigonometric signs sin, cos, and tan of a right-angled triangle.
- The right-angled triangle has three edges, the hypotenuse, the perpendicular, and the base.
- The six trigonometric ratios are sine, cosine, tangent, cotangent, cosecant, and secant.
- The ratios are expressed as Sin x = perpendicular/hypotenuse; Cos x = base/hypotenuse; Tan x = perpendicular/base; Cosec x = 1 / sin; Sec x = 1 / cos; Cot X = 1 / tan.
**The formula for cos square theta is, Cos**^{2}x = 1 – sin^{2}.- Other formulas of cos square theta are: Cos square x = cos 2x + 1; Cos square x = cos 2x + sin square x.

**Solved Examples using Cos Square Theta Formula**

**Example 1**: What is the value of cos square x, if Sin x = 4/5 ?

**Solution**:

Using Cos Square theta formula,

**Cos ^{2} x = 1 – Sin^{2} x**

= 1 – (4/5)^{2}

= 1 – 16/25

= (25 – 16) / 25

= 9/25

Thus, cos x = 3/5

**Example 2: **If cos^{2}x – sin^{2}x = 41/841, then find the value of cos^{2}x.

**Solution**:

Given: cos^{2}x – sin^{2}x = 41/841

We know that,

sin^{2}x = 1 – cos^{2}x

Substituting in the above equation we get,

cos^{2}x – (1 – cos^{2}x) = 41/841

⇒2 cos^{2}x – 1 = 41/841

⇒2 cos^{2}x = 1 + 41/841

⇒2 cos^{2}x = 882/841

⇒cos^{2}x = 882/(841 × 2)

⇒cos^{2}x = 441/841

Thus, the value of cos^{2}x is 441/841.

**Example: Determine the value of cosθ, with the value of sinθ given as 3/5.**

**Ans: **The value of sinθ is given as = 3/5

Now, by using the cos square formula, we can obtain:

⇒ cos^{2}θ + sin^{2}θ = 1

⇒ cos^{2}θ = 1 – sin^{2}θ = 1 – (3/5)^{2} = 1 – 9/25

⇒ cos^{2}θ = 16/25

⇒ cosθ = √(16/25) = ± 4/5

Thus, the value of cosθ is **± 4/5**.

**Example: Determine the value of cos2θ, with the value of cosθ given as = 1/2.**

**Ans: **By applying a generalized formula,

⇒ **cos2θ = 2cos ^{2}θ – 1**

The value of cos2θ, after substituting the value is, cosθ = 1/2

Hence,

⇒ cos2θ = 2 × (1/2)^{2} – 1 = 2/4 – 1 = 1/2 – 1 = – 1/2

Accordingly, **cos2θ = – 1/2.**

**Ques: What do you mean by trigonometric ratios?**

**Ans: **Trigonometric ratios are the value of the trigonometric functions that are based on the ratio of the primary trigonometric signs sin, cos, and tan of a right-angled triangle. They are used widely in mathematics for solving trigonometry and calculus problems.

**Ques: What are the different types of trigonometric ratios?**

**Ans: **There are six trigonometric ratios which are,

- Sine or sin
- Cosine or cos
- Tangent or tan
- Cosecant or cosec
- Secant or sec
- Cotangent or cot

**Ques: What are the values of the different trigonometric ratios?**

**Ans:** The trigonometry ratios are expressed as follows:

- Sine = perpendicular/hypotenuse
- Cosine = base/hypotenuse
- Tangent = perpendicular/base
- Cosecant = 1/sine = hypotenuse/perpendicular
- Secant = 1/cosine = hypotenuse/base
- Cotangent = 1/tangent = base/perpendicular

**Ques: What is the formula for cos square theta?**

**Ans: **The formula for cos square theta is derived from the following equation:

Cos^{2} x + sin^{2} x = 1

So we will get the formula for cos square x by shifting sin square x to the right hand side.

Cos^{2} x = 1 – sin^{2} x

Other formulas for cos square x are:

Cos^{2} x = cos 2x + 1

Cos^{2} x = cos 2x + sin^{2} x

**Ques: Let there be an angle with sin x value as 2/3. Calculate the value of cos x.**

**Ans: **As we know,

Sin square x + cos square x = 1

(2/3)^{2} + cos^{2} x = 1

Cos^{2} x = 1- (2/3)^{2}

Cos^{2} x = 1- 4/9

= (9-4)/9

= 5/9

Cos x = 5/3.

**Ques: The value of cos x is ¾ . Calculate the value of sin x.**

**Ans:** As we know,

Cos square x + sin square x = 1

Sin^{2} x = 1- cos^{2} x

= 1- (3/4)^{2}

= 1- 9/16

= (16-9)/16

= 7/16

Sin x = 7/4 .

**Ques: Derive the value of cos2x, if cos square x is 2/6.**

**Ans:** As we know,

Cos 2x = 2 cos^{2}2 x – 1

Cos 2x = 2 (2/6)^{2} – 1

Cos 2x = 2 (4/36) – 1

Cos 2x = (2/9) – 1

Cos 2x = (2-9)/1

Cos 2x = -7.

**Ques: Determine the value of cosθ, with the value given as, cosθ – sinθ = 1.**

**Ans: **The given value is, cosθ – sinθ = 1.

or, cosθ = 1 + sinθ —- (i)

By applying the cos square formula, we get

cos^{2}θ + sin^{2}θ = 1

cos^{2}θ = 1 – sin^{2}θ = (1 + sinθ)(1 – sinθ)

cos^{2}θ = cosθ (1 – sinθ)

cosθ (cosθ – 1 + sinθ) = 0

So, can obtain the two cases,

**cosθ = 0**

else, cosθ – 1 + sinθ = 0

or, cosθ = 1 – sinθ —- (ii)

From eq.(i) and eq.(ii), we get

1 – sinθ = 1 + sinθ

2sinθ = 0

sinθ = –

From eq.(i), we get cosθ = 1 + sinθ = 1 + 0 = 1

Thus, **cosθ = 1**

So, we get two possibilities. The value of cosθ is **0 **or** 1**.

**Ques: If cosθ = 3/5, then what is the value of sin ^{2}θ – cos^{2}θ?**

**Ans:** As per the question, the value of cosθ = 3/5

Now, using the cos square formula, we can express:

⇒ sin^{2}θ – cos^{2}θ = (1 – cos^{2}θ) – cos^{2}θ = 1 – 2cos^{2}θ

After we put the value of cosθ = 3/5, we can obtain:

⇒ sin^{2}θ – cos^{2}θ = 1 – 2cos^{2}θ

= 1 – 2 × (3/5)^{2}

= 1 – 2 × 9/25

= 1 – 18/25

= **7/25**

**Ques: What is the proof of cos ^{2}θ + sin^{2}θ = 1?**

**Ans: **The trigonometric functions for any right-angled triangle can be expressed as:

**cosθ = base/hypotenuse****sinθ = altitude/hypotenuse**

So, we can represent the same as:

→ cos^{2}θ + sin^{2}θ = base^{2}/hypotenuse^{2} + altitude^{2}/hypotenuse^{2}

hence,

cos^{2}θ + sin^{2}θ = (base^{2} + altitude^{2})/hypotenuse^{2}

By using the Pythagoras theorem for right-angled triangle, we can obtain:

base^{2} + altitude^{2} = hypotenuse^{2}

Thus, we acquire:

**cos ^{2}θ + sin^{2}θ = 1**

**Question 1. Find the value of cosθ, given the value of sinθ, is 3/5.**

**Solution:**

Given, the value of sinθ = 3/5

Using cos square formula, we get

cos

^{2}θ + sin^{2}θ = 1cos

^{2}θ = 1 – sin^{2}θ = 1 – (3/5)^{2}= 1 – 9/25cos

^{2}θ = 16/25cosθ = √(16/25) = ± 4/5

Thus, the value of cosθ is

± 4/5.

**Question 2. Find the value of cosθ, given the value of cosθ – sinθ = 1**

**Solution:**

Given, cosθ – sinθ = 1.

or, cosθ = 1 + sinθ —- (i)

Using cos square formula, we get

cos

^{2}θ + sin^{2}θ = 1cos

^{2}θ = 1 – sin^{2}θ = (1 + sinθ)(1 – sinθ)cos

^{2}θ = cosθ (1 – sinθ)cosθ (cosθ – 1 + sinθ) = 0

So, we get two cases,

cosθ = 0else, cosθ – 1 + sinθ = 0

or, cosθ = 1 – sinθ —- (ii)

From eq.(i) and eq.(ii), we get

1 – sinθ = 1 + sinθ

2sinθ = 0

sinθ = 0

From eq.(i), we get cosθ = 1 + sinθ = 1 + 0 = 1

Thus,

cosθ = 1So, we get two possibilities. The value of cosθ is

0or1.

**Question 3. If cosθ = 3/5, find the value of sin ^{2}θ – cos^{2}θ.**

**Solution:**

Given, the value of cosθ = 3/5

Now, using cos square formula, we can write

sin

^{2}θ – cos^{2}θ = (1 – cos^{2}θ) – cos^{2}θ = 1 – 2cos^{2}θPutting the value of cosθ = 3/5, we get

sin

^{2}θ – cos^{2}θ = 1 – 2cos^{2}θ = 1 – 2 × (3/5)^{2}= 1 – 2 × 9/25 =1 – 18/25 =7/25So, the answer is

7/25.

**Question 4. Find the value of cos2θ, given the value of cosθ = 1/2.**

**Solution:**

Using the generalized formula,

cos2θ = 2cos^{2}θ – 1we can find the value of cos2θ, by substituting cosθ = 1/2

cos2θ = 2 × (1/2)

^{2}– 1 = 2/4 – 1 = 1/2 – 1 = – 1/2Thus,

cos2θ = – 1/2.

Question: What is the value of cos square x, if Sin x = ⅗?Solution:

Cos^{2}x = 1 – Sin^{2}x= 1 – (⅗)

^{2}= 1 – 9/25

= (25 – 9) / 25

= 16/25

Cos x = 4/5

**Question 2: If sin A = 7/25, then find cos ^{2}A and cos^{2}A – sin^{2}A.**

Solution:

Given,

sin A = 7/25

We know that,

cos^{2}A = 1 – sin^{2}A

= 1 – (7/25)^{2}

= 1 – 49/625

= (625 – 49)/625

= 576/625

Therefore, cos^{2}A = 576/625

Now,

cos^{2}A – sin^{2}A = (576/625) – (49/625)

= 527/625

**Question 3: If cos ^{2}x – sin^{2}x = 41/841, then find the value of cos^{2}x**.

Solution:

Given,

cos^{2}x – sin^{2}x = 41/841

Substituting sin^{2}x = 1 – cos^{2}x in the above equation,

cos^{2}x – (1 – cos^{2}x) = 41/841

2 cos^{2}x – 1 = 41/841

2 cos^{2}x = 1 + 41/841

2 cos^{2}x = 882/841

cos^{2}x = 882/(841 × 2)

cos^{2}x = 441/841

## FAQs (Frequently Asked Questions)

### 1. What are trigonometric ratios?

Trigonometric functions of triangles are also called trigonometric ratios-Sine, cosine, and tangent are the three basic trigonometric ratios, also Sine is abbreviated as sin, cosine as cos, and tangent as tan. Sine, cosine and tangent are also widely used in the world of trigonometric functions in Mathematics and physics. Let’s understand with the help of right triangles. We all know that one side of a right triangle is 90°. The bottom surface is called the adjacent surface. The vertical side is called the counter. The left side is called the hypotenuse.

### 2. Why is cosθ or cos2θ important in trigonometric functions?

All periodic and aperiodic functions have their own meaning in the world of trigonometry. Cosine is also a periodic function and has wide applications in Mathematics. We can understand the cosine function with the help of the unit circle. The easiest way to understand the cosine function. First, draw a unit circle on coordinates centred at the angle θ. The right side has the positive x-axis and the left side has the negative x-axis. The positive y-axis is (0, sineθ). The positive x-axis is (cosθ, 0). The hypotenuse is (cosθ, sinθ).

### 3. How are the trigonometric functions defined?

Cosine, sine and tangent are the main trigonometric functions. Different angles to get the value of the function are 00, 300, 600, 900, 1800, 3600. 1 cos theta squared is used in various trigonometry calculations. Trigonometric calculations can be performed using the six trigonometric ratios used in trigonometry. These trigonometric ratios can be used to solve or derive almost any trigonometric calculation. Sine, cosine, tangent, cotangent, cos, and cosec are some of the trigonometric functions, where sine and cosine are basic ratios. Tan, cot, and sec are derived functions. Several ratios exist for sine, cot and tan. Sinx = opposite/hypotenuse. Cosx = adjacent/hypotenuse. Tanx = sinx/cosx.

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